Photogravity
, or black body temperature, of the Sun (5777 K) is the temperature a black body of the same size must have to yield the same total emissive power.]] The relationship between Energy(E) and momentum(P) of photon is as follows. E = h\nu= hc/\lambda P = hk/2\pi = h/\lambda P = E/c Photo-gravitational mass of central body To find the total absolute Momentum of light radiated for an object from mass m, we have to take into account the surface area, A(in m2) of the astronomic object: : P = A j^{\star} /c = A \varepsilon\sigma T_s^{4}/c. where j'' is known variously as the black-body '''irradiance', energy flux density, radiant flux, or the emissive power and k is proportional constant. The constant of proportionality σ, called the Stefan–Boltzmann constant or Stefan's constant, is non-fundamental in the sense that it derives from other known constants of nature. The value of the constant is : \sigma=\frac{2\pi^5 k^4}{15c^2h^3}= 5.670 400 \times 10^{-8} \textrm{J\,s}^{-1}\textrm{m}^{-2}\textrm{K}^{-4} where k is the Boltzmann constant, h is Planck's constant, and c is the speed of light in a vacuum. Thus at 100 K the energy flux density is 5.67 W/m2, at 1000 K 56,700 W/m2, etc. Define Photogavitational mass of central body as follows, m_{pg} = k_{pg} T_s^4 Where m_i is inertial mass and k_{pg} is photogravitational constant. However without bulk, the surface temperature couldn't be maintained. With T cube inertial mass assumption, T^4 = k_x m_i T_s and Photo-gravitational mass of centrifugal body can be written as follows. m_{pg} = m_i T_s /T_0 Thus Momentum from unit area P/A =k_p M_i T_s The difference between surface temperature and core temperature is also important for near field gravity. with lattitude, it increases resulting in value between 9.78 and 9.82 m/s2 thumb|[[gravitational mass density and temperature of structure of the Earth]] Attenuation by the space Unit time areal density of momentum is normally dependent on Inverse square of distance for perfect vacuum without any absorption, scattering and reflection by dark matter. Optical attenuation by Solid is described by the factor of exp(-kr). Although 1/rr factor is abnormal for opotics. density variation of dark matter from the astronomic object isn't negligible and is proportianl to -ln(rr)/r which is primitive. Function -2lnr/r has a minimum about -0.74 at r=2.73 and asymtotically approaches to 0. Density function inbetween the astronomical objects may be rewritten. D®/D_0 = C_l - 2 lnr/r where C_l > 0.74 and r is also normalized value of distance r=d/ r_0 To find r_0 & D_0 ; out is very necessary to proceed. Temperature and mass density against altitude from the NRLMSISE-00 standard atmosphere model shows that local minimum is between 100km and 160km. File:LunarPhotons.png| APOLLO Collaboration photon pulse return times File:Darkmatter.JPG|density of dark matter which gives 1/rr attenuation Image:Atmospheric_pressure.JPG|Atmospheric pressure Image:Atmospheric_density.JPG|Atmospheric density Image:Atmosphere model.png|Temperature and mass density against altitude from the NRLMSISE-00 standard atmosphere model Gravity Photogravitational force might be deduced as follows. F = \phi_1 (T_{s1} T_{s2}/r_{12})^4 where phi is photogravitaional constant and Ts is surface temperature of astronomic object. To take care of differential effect, It can be modified as follows. F = \phi_2 (\nabla_{s1} T \nabla_{s2} T/r_{12})^4 if Electrogravitational emission of centripetal body stimulated by photogravity of the central body might be proportional to mass rather than Ts biquadrate. We modify the above equation as follows, F = \phi_3 m_{2}(T_{s1}/r_{12})^4 Then, inverse biquadrate force is F = \phi_4 (T_{s1}/T_0) m_{1} m_{2} /r_{12}^4 Sunlight intensity in the Solar System Different bodies of the Solar System receive light of an intensity inversely proportional to the square of their distance from Sun. A rough table comparing the amount of light received by each planet on the Solar System follows (from data in http://www.starhop.com/High/SolInt-19.pdf): The actual brightness of sunlight that would be observed at the surface depends also on the presence and composition of an atmosphere. For example Venus' thick atmosphere reflects more than 60% of the solar light it receives. The actual illumination of the surface is about 5,000–10,000 lux, comparable to that of Earth during a dark, very cloudy day. Sunlight on Mars would be more or less like daylight on Earth wearing sunglasses, and as can be seen in the pictures taken by the rovers, there is enough diffuse sky radiation that shadows would not seem particularly dark. Thus it would give perceptions and "feel" very much like Earth daylight. For comparison purposes, sunlight on Saturn is somewhat slightly brighter than Earth sunlight on the average sunset or sunrise. Even on Pluto the Sun would be still bright enough to almost match the average living room. To see the Sun shine as dim as the full Moon on the Earth, a distance of about 500 AU (~69 light-hours) is needed: there is only a handful of objects in the solar system known to orbit farther than such a distance, among them 90377 Sedna and (87269) 2000 OO67. See also *Stefan–Boltzmann law *AME *Mass *Esther -Electra *Gravitational mass *sunlight Category:gravity wikia Category:photogravity